A Space is Reflexive if its Image under the Canonical Injection is Reflexive?

Question

Consider the following corollary in Brezis: Here is part of the proof of it:

It was mentioned that if $J(E) \subseteq E^{**}$ is reflexive, then $E$ is also reflexive, where $J: E \to E^{**}$ be the canonical injection. Why is this true?

Answer 1

This reduces to prove the following claim: Let $E$ and $F$ be Banach spaces. If $T:E \to F$ is a surjective isometry, then $E$ is reflexive if and only if $F$ is reflexive. We show only one direction as the other way is symmetric. Since $T$ is surjective and an isometry, it must be bijective as isometry implies injectivity. This implies that the adjoint operator $T^*: F^* \to E^*$ is also bijective as $(T^*)^{-1} = (T^{-1})^*$. Similarly we have $T^{**}: E^{**} \to F^{**}$ to be a bijection as well. Now we claim that the canonical injection $$ J_F: F \to F^{**} $$ can be written as $J_F \equiv T^{**} \circ J_E \circ T^{-1}$, where $J_E: E \to E^{**}$ is the canonical injection (in this case is bijective as $E$ is reflexive). Granted this is true, then $J_F$ is a bijection, so indeed a surjective mapping. Now to show the claim, we need to show $$ T^{**} \circ J_E \circ T^{-1}(f) = J_F(f) $$ for all $f \in F$. To show this, fix $f \in F$, we need to show $$ T^{**} \circ J_E \circ T^{-1}(f)(x) = J_F(f)(x) $$ for all $x \in F^{*}$. Now fix $x \in F^{*}$, we have \begin{align} T^{**} \circ J_E \circ T^{-1}(f)(x) &= (T^{**} \circ J_E \circ T^{-1}(f), x)\\ &= (J_E \circ T^{-1}(f), T^*x)\\ &= (T^*x, T^{-1}(f)) \\ &= (x, T T^{-1}(f))\\ &= (x, f) = J_F(f)(x). \end{align} This shows that we indeed have $J_F \equiv T^{**} \circ J_E \circ T^{-1}$.